Question

By how many

- {CH}_{2} -

(methylene) units do the formulae of the first two members of homologous series of alkanes, methane

({CH}_{4})

and ethane

(C_{2} H_{6})

differ? Similarly, by how many

- {CH}_{2} -

units do the neighbouring members ethane

(C_{2} H_{6})

and propane

(C_{3} H_{8})

differ from each other?

Accepted Answer

A homologous series is a series that has compounds with the same functional group and each member is differed by $- {CH}_{2} -$ group. A general formula is used to represent each homologous series. The alkane series is also a homologous series with the general formula $C_{n} H_{2 n + 2}$ and each member will differ by a $- {CH}_{2} -$ group.

The first member of the alkane series is methane with the formula $(C_{2} H_{6})$ and by adding a ${CH}_{2}$ group we will get the second member of the alkane series which is ethane, $\underset{Methane}{{CH}_{4}} + \underset{methylene}{{CH}_{2}} \Rightarrow \underset{Ethane}{C_{2} H_{6}}$ . Similarly adding a ${CH}_{2}$ group to ethane will give a propane molecule with the formula $C_{3} H_{8}$ , $\underset{Ethane}{C_{2} H_{6}} + \underset{methylene}{{CH}_{2}} \Rightarrow \underset{propane}{C_{3} H_{8}}$ .

Maharashtra Board Solutions for Chapter: Carbon Compounds, Exercise 10: Use your brain power!

Maharashtra Board Science and Technology Solutions for Exercise - Maharashtra Board Solutions for Chapter: Carbon Compounds, Exercise 10: Use your brain power!

Questions from Maharashtra Board Solutions for Chapter: Carbon Compounds, Exercise 10: Use your brain power! with Hints & Solutions

Learn and Prepare for any exam you want !